how to find ionisation potential

how to find ionisation potential



2 Answers

33 Points
11 years ago

Actually ionisation potential means that "the amount of enrgy required to remove the electron from the valency shell of a particular atom".We have to find it as the depend upon its atomic size.It means that low size atoms have a low tendency to removing the electron.

chinnababu allam
29 Points
10 years ago

The Ionization Potentials and Shielding
 of Electrons in the First Shell

The Bohr model of a hydrogen-like ion predicts that the total energy E of an electron is given by

E = −Z²R/n²

where Z is the net charge experienced by the electron, n is the principal quantum number and R is a constant
equal to approximately 13.6 electron volts (eV).  This formula is the result of the total energy being
equal to

E = − Ze²/(2r


where e is the charge of the electron and r
n is the orbit radius when the principal quantum
number is n.  The orbit radius is given by


 = n²h/(Zm


where h is Planck''s constant divided by 2π and m
e is the mass of the electron.

Shell Structure

Electrons in atoms are organized in shells whose capacities are equal to 2m², where m is an integer.  Thus there
can be at most 2 electrons in the first shell,  8 in the second shell and 8 in the third shell and 18 in each of the fourth
and fifth shells.  Here only the first shell
is being considered.  For elements above helium all of the electrons have been removed except two.

Here are all the ionization potentials for such ions.

The Ionization Potentials for the

in the First Shell for the

 Elements Helium through Copper











2    54.41778    24.58741

3    122.45429    75.64018

4    217.71865    153.89661

5    340.2258    259.37521

6    489.99334    392.087

7    667.046        552.0718

8    871.4101    739.29

9    1103.1176    953.9112

10    1362.1995    1195.8286

11    1648.702    1465.121

12    1962.665    1761.805

13    2304.141    2085.98

14    2673.182    2437.63

15    3069.842    2816.91

16    3494.1892    3223.78

17    3946.296    3658.521

18    4426.2296    4120.8857

19    4934.046    4610.8

20    5469.864    5128.8

21    6033.712    5674.8

22    6625.82    6249

23    7246.12    6851.3

24    7894.81    7481.7

25    8571.94    8140.6

26    9277.69    8828

27    10012.12    9544.1

28    10775.4    10288.8

29    11567.617    11062.38

The Ionization Potential of the First Electron

as a Function of Proton Number

An ion with only one electron is equivalent to the hydrogen atom but having a positive
charge of Z instead of one.  The Bohr theory applies to such system.  According to the
Bohr theory the ionization potential should be

IE = Z²R/n²

R is constant and n, the quantum number, is equal to 1.  Thus the ionization potential should be proportional
to the proton number squared.  Here is the plot of the relationship.

It certainly looks like a quadratic relationship.  To test this proposition more precisely the logarithms of the
ionization potential and the proton numbers are computed.  The plot of these quantitites is shown below.

The regression equation of the logartithm of ionization potential on the logarithm of the proton number is

ln(IE) = 2.604026184 + 2.004055992ln(#p)

[2029.1] [4123.2]

 R² = 0.999998471

The numbers in the square brackets are the t-ratios for the regression coefficients.  For a regression coefficient to
be statistically significantly different its magnitude must be greater than 2.0.  As can be seen the regression coefficient for
ln(#p) is highly significant.  And the value appears to be essentially 2, as the Bohr theory predicts. However this proposition
is tested by comparing the difference between the regression coefficient and 2 with the standard deviation of the regression
coefficient.  Although the difference is small, 0.004056, the standard deviation of the regression coefficient is even smaller,
0.000486039, the ratio is 8.3.  Common sense however says that the Bohr theory is verified for the first electron in the
the first shell.

The Z in the formula is the net charge experienced by the electron, the number of protons #p in the nucleus less any
shielding ε. The Bohr theory equation is then

IE = R(#p−ε)²

which can be expressed as

IE = R(#p² − 2#p*ε + ε²)

where R is the Rydberg constant, 13.6 eV.

 Thus the appropriate regression equation would be

IE = c


 + c


#p + c



in which






 ≅ R

Since in this case there seems to be no reason for there to be any shielding for the first electron the value of ε should be essentially zero.

The results of the regression are:

IE = 17.63611461 − 4.994593419#p + 13.89254281(#p)²

[5.1] [-9.7] [858.9]

R² = 0.999998221

The coefficient of #p is negative but statistically different from zero. The coefficient of (#p)² is notably close
to 13.6.  The value of ε can be found as

ε = ½(−c




) = 0.179758072

Somehow some small portion of the charge of the nucleus is being shielded.


The Bohr theory works marvelously well for an atom or ion with one electron but not nearly so well for multiple electrons.

One line of approach is to treat the atom or ion with two electrons as a many-body problem.  This approach does not come up
with much in the way of definite results.  An alternate line of approach is to say that electrons in the same shell shield
some fraction of a positive charge in the nucleus.  This

shielding ratio would be based upon the proportion of the
charge of an electron which is closer to the nucleus than the center of an electron.  The value of one half would be
plausible.  However the shielding ratio could be affected by any deviation from spherical symmetry.

The Shielding Ratios of the Electrons in the First Shell

as a Function of Proton Number of the Nucleus

The shielding ratio ρ for the electrons in the first shell can be computed as

ρ = 1 − E





where E
1 and E
2 are the ionization potentials for the first and second electrons, respectively.

The values for the elements for which the data is available in the CRC Handbook of Physics and Chemistry
82nd Edition (2001-2002)

 are given below.

The Shielding Ratio of

 the Electrons
in the First Shell

for the Elements Helium through Copper





2    0.548173226

3    0.382298652

4    0.293139977

5    0.237638033

6    0.199811573

7    0.172363225

8    0.151616443

9    0.135258834

10    0.122134019

11    0.111348807

12    0.10234044

13    0.09468214

14    0.088116709

15    0.082392514

16    0.077388254

17    0.072922812

18    0.068985102

19    0.065513374

20    0.062353287

21    0.059484443

22    0.056871451

23    0.054487091

24    0.052326782

25    0.050319998

26    0.048470039

27    0.046745345

28    0.045158416

29    0.043676844

The graph of these data displays a very regular pattern.

This appears to be a relationship of the form

ρ = α(#p)-β

The value of the parameters α and β can be found by plotting the logarithms of ρ and #p, as below.

A regression of ln(ρ) on ln(#p) gives:

ln(ρ) = 0.093012504 − 0.956463724ln(#p)

[16.2] [-439.2]

R² = 0.999865247

The results indicate that the shielding ratio is predictable on the basis of the proton number, but the shielding of positive charge in the nucleus by the electrons in the first shell
is primarily important for small nuclides.

The Ionization Potential for the Second

 Electron as a Function of the Proton Number

Previously a regression equation of the form

IE = c


 + c


#p + c



was applied to the data for the first electron and it was found that there was a shielding of 0.179758072 units of charge.
When the data for the ionization potentials of the second electron is used the regression results are

IE = 20.41371337 − 21.54911763#p + 13.85998758(#p)²

 [6.2] [-44.6] [912.9]

R² = 0.999998302

Again the coefficient of #p is negative and the coefficient of (#p)² is notably close to the Rydberg constant of 13.6.
The value of the shielding that is found from the coefficients is 0.777385892.  The difference between this value and the
value for the first electron is 0.59762782, indicating that the second electron shields about 0.6 of a unit charge of the
nucleus.  This is notably close to the value of 0.5 suggested by the simplest formulation of the notion of shielding by
electrons in the same shell.

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